Question

The frequency of vibrations of a piano string varies directly as the square root of the tension on the string and inversely as the length of the string. The middle A string has a frequency of 440 vibrations per second. Find the frequency of a string that has 1.25 times as much tension and is 1.2 times as long.

Answer 1

were you able to derive an equation based on the information supplied ?

Answer 2

No, I am terrible at word problems

Answer 3

let f = frequency

we know it is directly proportional (varies directly) as the square root of tension. we can call that \[\sqrt{T}\]

Frequency also varies inversely (varies indirectly) as the length of the string. we can call that \[L\]

\[f = \frac{ k \sqrt{T} }{ L }\]

where k is the constant of proportionality. will this information help ?

Answer 4

Could you work it with me

Answer 5

try deriving an equation for k using the fact that the frequency is 440 vibrations per second.

Answer 6

So would it be 440=k(square root 1.25) divided by 1.2

Answer 7

\[f_{A} = \frac{ k \sqrt{T_{A}} }{ L_{A} } = 440 \rightarrow k = \frac{ 440 L_{A} }{ \sqrt{T_{A}} }\]

The new string has a tension 1.25 as much as Ta and length 1.2 x as much, so our new string as a frequency . . .

\[f_{S} = \frac{ k \sqrt{T _{S}}}{ L _{S} }\]

We know Ts = 1.25 Ta, and Ls = 1.2 La, and k is given above. so finally:

\[f_{S} = \frac{ 440 L_{A} }{ \sqrt{T_{A}} } * \frac{ \sqrt{1.25 T_{A}} }{ 1.2 L_{A} }\]

cancel the units and obtain the desired result

Answer 8

so would my answer be fs =410

Answer 9

that sounds reasonable

Answer 10

and the calculator agrees

Answer 11

is it 410 vibrations per second

Answer 12

yep. i get 409.946... so 410

Answer 13

Thank you